Consider the maps \(g, h: S^1 \to S^1\) given by \(g(z) = z^n\) and \(h(z) = 1/z^n\). (Here we represent \(S^1\) as the set of complex numbers \(z\) of absolute value 1.) Compute the induced homomorphisms \(g_*\), \(h_*\) of the infinite cyclic group \(\pi_1(S^1, b_0)\) into itself. [Hint: Recall the equation \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta\).]
Solution
1. Identification of \(\pi_1(S^1, 1)\) with \(\mathbb{Z}\):Proof:
1.1. Take the basepoint \(b_0 = 1 = e^{i \cdot 0} \in S^1\).
1.2. The standard fundamental loop generating \(\pi_1(S^1, 1) \cong \mathbb{Z}\) is:\[\gamma(s) = e^{2\pi i s} = \cos(2\pi s) + i \sin(2\pi s) \quad \text{for } s \in [0, 1].\]
1.3. Under the standard covering map \(p: \mathbb{R} \to S^1\) with \(p(t) = e^{2\pi i t}\), the unique lift \(\tilde{\gamma}: [0, 1] \to \mathbb{R}\) starting at \(0\) is \(\tilde{\gamma}(s) = s\), with terminal point \(\tilde{\gamma}(1) = 1\).
1.4. Thus the isomorphism \(\operatorname{deg}: \pi_1(S^1, 1) \xrightarrow{\cong} \mathbb{Z}\) sends \([\gamma] \mapsto 1\).
2. Computation of \(g_*\):Proof:
2.1. The composition \(g \circ \gamma: [0, 1] \to S^1\) is:\[(g \circ \gamma)(s) = g(e^{2\pi i s}) = (e^{2\pi i s})^n = e^{2\pi i n s} = \cos(2\pi n s) + i \sin(2\pi n s).\]
2.2. The unique lift of \(g \circ \gamma\) to \(\mathbb{R}\) starting at \(0\) is \(\widetilde{g \circ \gamma}(s) = n s\).
2.3. The endpoint is \(\widetilde{g \circ \gamma}(1) = n\).
2.4. Thus \(g_*([\gamma]) = [g \circ \gamma] = [\gamma]^n\).
2.5. Expressing \(\pi_1(S^1, 1) \cong \mathbb{Z}\) additively, \(g_*\) is multiplication by \(n\):\[g_*(k) = n k \quad \text{for all } k \in \mathbb{Z}.\]
3. Computation of \(h_*\):Proof:
3.1. The composition \(h \circ \gamma: [0, 1] \to S^1\) is:\[(h \circ \gamma)(s) = h(e^{2\pi i s}) = (e^{2\pi i s})^{-n} = e^{-2\pi i n s} = \cos(-2\pi n s) + i \sin(-2\pi n s).\]
3.2. The unique lift of \(h \circ \gamma\) to \(\mathbb{R}\) starting at \(0\) is \(\widetilde{h \circ \gamma}(s) = -n s\).
3.3. The endpoint is \(\widetilde{h \circ \gamma}(1) = -n\).
3.4. Thus \(h_*([\gamma]) = [h \circ \gamma] = [\gamma]^{-n}\).
3.5. Expressing \(\pi_1(S^1, 1) \cong \mathbb{Z}\) additively, \(h_*\) is multiplication by \(-n\):\[h_*(k) = -n k \quad \text{for all } k \in \mathbb{Z}.\]
4. Q.E.D.By steps 2 and 3, under the identification \(\pi_1(S^1, 1) \cong \mathbb{Z}\) of step 1, \(\boxed{g_*(k) = n k \text{ and } h_*(k) = -n k}\).